Factor completely. $(x^2-5x+4)(x^2-9)=$
Explanation: The expression is already somewhat factored, because it is given as the product of two factors, $(x^2-5x+4)$ and $(x^2-9)$. To completely factor the expression, we need to factor each of these factors further. Factoring $(x^2-5x+4)$ $x^2-5x+4=(x-4)(x-1)$ Factoring $(x^2-9)$ We notice this expression has the difference of squares pattern: $\begin{aligned} &\phantom{=}x^2-9 \\\\ &=(x)^2-(3)^2 \\\\ &=(x+3)(x-3) \end{aligned}$ Putting it all together $\begin{aligned} &\phantom{=}{(x^2-5x+4)}C{(x^2-9)} \\\\ &={(x-4)(x-1)}C{(x+3)(x-3)} \end{aligned}$ In conclusion, this is the completely factored expression: $(x-4)(x-1)(x+3)(x-3)$